The Simplex Algorithm
LI Xiao-lei
preface
• The two-variable LP program can be solved
graphically.
• Most real-life LPs have many variables.
• In many industrial applications, the simplex
algorithm is used to solve LPs with
thousands of constraints and variables.
How to convert an LP to
standard form
• Before the simplex algorithm can be
used to solve an LP, the LP must be
converted into an equivalent problem
in which all constraints are equations
and all variables are nonnegative. An
LP in this form is said to be in
standard form.
EXAMPLE 1
• Leather limited manufactures two types of
belts: the deluxe model and the regular
model. Each type requires 1 sq yd of
leather. A regular belt requires 1 hour of
skilled labor, and a deluxe belt requires 2
hours. Each week, 40 sq yd of leather and
60 hours of skilled labor are available.
Each regular belt contributes $3 to profit
and each deluxe belt, $4.
EXAMPLE 1
• Define,
X1=number of deluxe belts produced weekly
X2=number of regular belts produced weekly
• The LP is,
Max z=4x1+3x2
s.t.
x1+ x2≤40
2x1+ x2≤60
x1,x2≥0
(LP1)
(Leather constraint)
(Labor constraint)
Convert inequality constraints
to equality constraints
• Define for each ≤ constraint a slack
variable si (si=slack variable for ith
constraint).
We define,
s1=40- x1-x2 or x1+ x2+ s1=40
s2=60-2x1-x2 or 2x1+ x2 + s2=60
Note: a point satisfies the ith constraint if and
only if si≥0.
LP1 in standard form
Max z=4x1+3x2
s.t.
x1+ x2 +S1
=40
2x1+ x2
+S2 = 60
x1,x2,S1,S2≥0
(LP1’)
Note: if constraint i of an LP is a ≤ constraint, we
convert it to an equality constraint by adding a
slack variable si to the ith constraint and
adding the sign restriction si ≥0.
• Similarly, to convert the ith ≥ constraint to
an equality constraint, we define an excess
variable (some times called a surplus
variable) ei.
Note: if the ith constraint of an LP is a ≥
constraint, it can be converted to an equality
constraint by subtracting an excess variable ei
from the ith constraint and adding the sign
restriction ei≥0.
example
Original LP,
max z=20x1 + 15x2
s.t.
x1
≤100
x2 ≤100
50x1+ 35x2≤6000
20x1+ 15x2≥2000
x1,x2≥0
example
Standard form LP,
max z=20x1 + 15x2
s.t.
x1
+ s1
=100
x2
+s2
=100
50x1+ 35x2
+s3 =6000
20x1+ 15x2
-e4=2000
xi≥0(i=1,2); s3≥0(i=1,2,3); e4≥0
Preview of the simplex algorithm
• Suppose an LP with m constraints in
standard form,
max z= c1x1+ c2x2+…+ cnxn
s.t.
a11x1+a12x2+…+a1nxn=b1
a21x1+a22x2+…+a2nxn=b2
…
…
am1x1+am2x2+…+amnxn=bm
xi≥0(i=1,2,…,n)
(1)
• We define,
 a11
a
A   21
 

am1
a12
a22

am 2
 a1n 
 x1 
 b1 
x 
b 
 a2 n 
x   2 b   2 


 

 
 
 amn 
 xn 
bm 
The constraints may be written as the system
of equations Ax=b
Basic and nonbasic variables
• Consider a system Ax=b of m linear
equations in n variables (assume n≥m)
• DEFINATION
A basic solution to Ax=b is obtained by setting nm variables equal to 0 and solving for the
values of the remaining m variables. This
assumes that setting the n-m variables equal to
0 yields unique values for the remaining m
variables or, equivalently, the columns for the
remaining m variables are linearly independent.
Basic and nonbasic variables
• To find a basic solution to Ax=b, we
choose a set of n-m variables (the
nonbasic variables, or NBV) and set
each of these variables equal to 0.
Then solve for the values of the
remaining n-(n-m)=m variables (the
basic variables, or BV) that satisfy
Ax=b.
Basic and nonbasic variables
• The different choices of nonbasic
variables will lead to different basic
solutions.
To illustrate,
x1+x2
=3
-x2+x3=-1
We begin by choosing a set of 3-2=1 nonbasic
variables. If NBV={x3} then BV={x1,x2}. We find
that x1=2, x2=1, by setting x3=0. If NBV={x2}
then we get x1=3, x2=0, x3=-1.
Basic and nonbasic variables
• Some sets of m variables do not yield a
basic solution.
For example,
x1+2x2+x3=1
2x1+4x2+x3=3
If we choose NBV={x3}, then
x1+2x2=1
2x1+4x2=3
The system has no solution, there is no basic
solution to BV={x1,x2}
Feasible solutions
• DEFINATION
Any basic solution to (1) in which all variables are
nonnegative is a basic feasible solution(or bfs).
THEOREM 1
The feasible region for any linear programming problem is
a convex set. Also, if an LP has an optimal solution, there
must be an extreme point of the feasible region that is
optimal.
THEOREM 2
For any LP, there is a unique extreme point of the LP’s
feasible region corresponding to each basic feasible
solution. Also, there is at least one bfs corresponding to
each extreme point of the feasible region.
Feasible solutions
Illustration of Leather Limited example,
Max z=4x1+3x2
s.t.
x1+ x2 +S1
=40
2x1+ x2
+S2 = 60
x1,x2,S1,S2≥0
(LP1’)
60
D
50
B
x
2
40
30
E
20
10
0
C
0
10
20
30
x1
Basic
variables
Nonbasic
variables
x1,x2
x1,s1
x1,s2
x2,s1
x2,s2
s1,s2
s1,s2
x2,s2
x2,s1
x1,s2
x1,s1
x1,x2
A
40
50
60
Basic feasible solution
Corresponds
to corner
point
s1=s2=0, x1=x2=20
E
x2=s2=0,x1=30,s1=10
C
x2=s1=0,x1=40,s2=-20 Not a bfs
x1=s2=0,s1=-20,x2=60 Not a bfs
x1=s1=0,x2=40,s2=20 B
x1=x2=0,s1=40,s2=60 F
Adjacent basic feasible
solutions
• DEFINATION
For any LP with m constraints, two basic
feasible solutions are said to be
adjacent if their sets of basic variables
have m-1 basic variables in common.
How the simplex algorithm
solves LPs in a max problem
• Step1
find a bfs to the LP. We call this bfs the initial basic
feasible solution. In general, the most recent bfs will be
called the current bfs, so at the beginning of the
problem the initial bfs is the current bfs.
• Step2
determine if the current bfs is an optimal solution to the
LP. If it is not, find an adjacent bfs that has a larger zvalue.
• Step3
Return to step2, using the new bfs as the current bfs.
How the simplex algorithm
solves LPs in a max problem
• If an LP in standard form has m constraints
and n variables, there may be a basic
solution for each choice of nonbasic
variables. From n variables, a set of n-m
nonbasic variables can be chosen in
n 
n!
  
 m  (n  m)!m!
Different ways.
How the simplex algorithm
solves LPs in a max problem
n 
 
 m
• Thus, an LP can have at most
basic
solutions.
• If we were proceed from the current bfs
to a better bfs, we would surely find the
optimal bfs after a finite number of
calculations.
The Simplex Algorithm
• Step1 convert the LP to standard form.
• Step2 obtain a bfs (if possible) from the
standard form.
• Step3 determine whether the current bfs is
optimal.
• Step4 if the current bfs is not optimal, determine
which nonbasic variable should become a basic
variable and which basic variable should become a
nonbasic variable to find a new bfs with a better
objective function value.
• Step5 use ero(elementary row operation)’s to find
the new bfs with the better objective function
value. Go back to step3.
example
• The dakota furniture company manufactures desk, tables,
and chairs. The manufacture of each type of furniture
requires lumber and two types of skilled labor: finishing
and carpentry.
Resource
Desk
Table
Chair
Lumber
8 board ft
6 board ft
1 board ft
Finishing hours
4 hours
2 hours
1.5hours
1.5hours
0.5 hour
Carpentry hours 2 hours
48 board feet of lumber, 20 finishing hours, and 8 carpentry hours are
available. A desk sells for $60, a table for $30, and a chair for $20.
demand for desks and chairs is unlimited, but at most five table can be
sold. Dakota wants to maximize total revenue.
example
• Defining the decision variable as
x1=number of desks produced
x2=number of tables produced
x3=number of chairs produced
• Dakota is the following LP:
max z=60x1+30x2+ 20x3
s.t.
8x1+ 6x2+ x3≤48
4x1+ 2x2+1.5x3≤20
2x1+1.5x2+0.5x3≤8
x2
≤5
x1,x2,x3≥0
(lumber constraint)
(finishing constraint)
(carpentry constraint)
(limitation on table demand)
example
• Convert the LP to standard form
max z=60x1+30x2+ 20x3
s.t.
8x1+ 6x2+ x3 +s1
=48
4x1+ 2x2+1.5x3
+s2
=20
2x1+1.5x2+0.5x3
+s3 =8
x2
+s4=5
x1,x2,x3, s1 , s2, s3 , s4 ≥0
Note: the row 0 for the objective function is
z -60x1-30x2-20x3=0
example
• Putting rows 1-4 together with row 0 and the sign
restrictions yields the equations and basic
variables.
Canonical form 0
Row 0
Row 1
Row 2
Row 3
Row 4
z -60x1-30x2-20x3
=0
8x1+ 6x2+ x3 +s1
=48
4x1+ 2x2+1.5x3
+s2
=20
2x1+1.5x2+0.5x3
+s3 =8
x2
+s4=5
Basic
variable
z=0
s1=48
s2=20
s3=8
s4=5
example
• The feasible solution for the initial
canonical form has
BV={s1,s2,s3,s4} ={0,48,20,8,5}
and NBV={x1,x2,x3}={0,0,0}
The feasible solution,
z=0,s1=48,s2=20,s3=8,s4=5,x1=x2=x3=0
Note: a slack variable can be used as a basic
variable for an equation if the right-hand side
of the constraint is nonnegative.
example
• Is the current basic feasible solution
optimal?
To determine whether there is any way that z
can be increased by increasing some nonbasic
variables at their current values of zero.
Note: from z=60x1+30x2+ 20x3 we can observe that x1 has
the most negative coefficient in row 0. we call x1 the
entering variable.
example
• Determine the entering variable
We choose the entering variable (in a max
problem) to be the nonbasic variable
with the most negative coefficient in
row 0(ties may be broken in an arbitrary
fashion).
Increasing x1 may cause a basic variable
to become negative.
example
• How to increasing x1( holding x2=x3=0) ?
From row 1 s1≥0 for x1≤48/8=6
From row 2 s2≥0 for x1≤20/4=5
From row 3 s3≥0 for x1≤8/2=4
From row 4 s4≥0 for all values of x1
Note: to keep all the basic variables nonnegative, the
largest that we can make x1 is min{48/8,20/4,8/2}=4.
Also for any row in which the entering variable had a
positive coefficient, the row’s basic variable became
negative when the entering variable exceeded
right-hand side of row
coefficient of entering variable in row
(10)
example
• How to increasing x1( holding x2=x3=0) ?
The ratio test
When entering a variable into the basis, compute
the ratio in (10) for every constraint in which
the entering variable has a positive coefficient.
The constraint with the smallest ratio is called
the winner of the ratio test. The smallest
ratio is the largest value of the entering
variable that will keep all the current basic
variables nonnegative.
example
• Find a new basic feasible solution: pivot in the
entering variable
In which row does the entering variable become basic?
Always make the entering variable a basic variable in a
row that wins the ratio test.
In example, to make x1 a basic variable in row 3, we use
elementary row operations to make x1 have a
coefficient of 1 in row 3 and a coefficient of 0 in all
other rows. This procedure is called pivoting on row 3;
and row 3 is the pivot row. The final result is that x1
replaces s3 as the basic variable for row 3. The term
in the pivot row that involves the entering basic
variable is called the pivot term.
example
• To make x1 a basic variable in row 3 by performing the
following ero’s.
ero 1 create a coefficient of 1 for x1 in row 3 by multiplying row
3 by ½. The resulting row is
x1+0.75x2+0.25x3+0.5s3=4 (row 3’)
ero 2 to create a zero coefficient for x1 in row 0, replace row 0
with 60(row 3’) +row 0.
z+15x2-5x3+30s3=240
(row 0’)
ero 3 to create a zero coefficient for x1 in row 1, replace row 1
with -8(row 3’)+row 1.
-x3+s1-4s3=16
(row 1’)
ero 4 to create a zero coefficient for x1 in row 2, replace row 2
with -4(row 3’)+row 2.
-x2+0.5x3+s2-2s3=4
(row 2’)
example
• x1 does not appear in row 4. we don’t need
to perform an ero to eliminate x1 from row
4.
Canonical form 1
Row 0’
Row 1’
Row 2’
Row 3’
Row 4’
z +15x1
- 5x3
- x3 +s1
- x2+ 0.5x3 +s2
x1+0.75x2+0.25x3
x2
+30s3
- 4s3
- 2s3
+0.5s3
=240
=16
=4
=4
+s4=5
Basic
variable
z=240
s1=16
s2=4
x1=4
s4=5
example
• The feasible solution for the canonical form 1 has
BV={z,s1,s2,x1,s4} and NBV={s3,x2,x3}
The feasible solution,
z=240,s1=16,s2=4,x1=4,s4=5,x2=x3=s3=0
In obtaining canonical form 1 from the initial canonical
form, we have gone from one bfs to a better (larger zvalue) bfs.
Note: the initial bfs and the improved bfs are adjacent.
The procedure used to go from one bfs to a better
adjacent bfs is called an iteration (or sometimes, a pivot)
of the simplex algorithm.
example
• Try to find a bfs that has a still larger z-value.
Rearranging row 0’ to solve for z,
z=240-15x2+5x3-30s3
• Increasing x2 by 1 will decrease z by 15.
• Increasing s3 by 1 will decrease z by 30.
• Increasing x3 by 1 will increase z by 5.
Thus we choose to enter x3 into the basis.
Note: the rule for determining the entering variable is to
choose the variable with the most negative coefficient in
the current row 0. since x3 is the only variable with a
negative coefficient in row 0’, it should be entered into
the basis.
example
• To determine how large x3 can be.
– From row 1’: s1=16+x3
all values of x3
– From row 2’: s2=4-0.5x3 x3≤4/0.5=8
– From row 3’: x1=4-0.25x3 x3≤4/0.25=16
– From row 4’: s4=5
all values of x3
The largest we can make x3 is
min{4/0.5,4/0.25}=8
example
• To determine how large x3 can be.
Also can be discovered by using (10) and the
ratio test:
Row 1’: no ratio (x3 has negative coefficient in row)
Row 2’:4/0.5=8
Row 3’:4/0.25=16
Row 4’:no ratio (x3 has a nonpositive coefficient in row 4)
Thus, the smallest ratio occurs in row 2’, and row
2’ wins the ratio test.
example
• Use ero’s to make x3 a basic variable
ero 1 create a coefficient of 1 for x3 in row 2’ by replacing row 2’
with 2(row 2’)
-2x2+x3+2s2-4s3=8
(row 2’’)
ero 2 to create a zero coefficient for x3 in row 0’, replace row 0
with 5(row 2’’) +row 0’.
z+5x2+10x2+10s3=280
(row 0’’)
ero 3 to create a zero coefficient for x3 in row 1’, replace row 1’
with row 2’’+row 1’.
-2x2+s1+2s2-8s3=24
(row 1’’)
ero 4 to create a zero coefficient for x3 in row 3’, replace row 3’
with -1/4(row 2’’)+row 3’.
x1+1.25x2-0.5s2+1.5s3=2 (row 3’’)
example
• x3 already has a zero coefficient in row 4’. we don’t
need to perform an ero to eliminate x3 from row 4’.
Canonical form 2
Row 0’’
Row 1’’
Row 2’’
Row 3’’
Row 4’’
z
+5x2
+10s2 +10s3
=280
-2x2
+s1 +2s2 - 8s3
=24
-2x2+x3
+2s2 - 4s3
=8
x1+1.25x2-0.5s2
+1.5s3
=4
x2
+s4=5
Basic
variable
z=280
s1=24
x3=8
x1=4
s4=5
example
• From canonical form 2, we find
BV={z,s1,x3,x1,s4} and NBV={s2,s3,x2}
It yields the following bfs:
z=280,s1=24,x3=8,x1=2,s4=5,s2=s3=x2=0
Since the bfs’s for canonical forms 1 and 2 have
4-1=3 basic variables in common (s1,s4,x1), they
are adjacent basic feasible solutions.
Now the second iteration (or pivot) has been
completed.
example
• If we rearrange row 0’’ and solve for z, we
obtain
z=280-5x2-10s2-10s3
We see that increasing any nonbasic variable will
cause z to decrease. this might lead us to
believe that our current bfs is an optimal
solution.
• Is a canonical form optimal(max problem)?
A canonical form is optimal( for a max problem)
if each nonbasic variable has a nonnegative
coefficient in the canonocal form’s row 0.
Summary of the simplex
algorithm for a max problem
• Step 1 convert the LP to standard form.
• Step 2 find a basic feasible solution.
• Step 3 if all nonbasic variables have
nonnegative coefficients in row 0, the
current bfs is optimal. If any variables in
row 0 have negative coefficients, choose
the variable with the most negative
coefficient in row 0 to enter the basis. We
call this variable the entering variable.
Summary of the simplex
algorithm for a max problem
• Step4 use ero’s to make the entering
variable the basic variable in any row
that wins the ratio test. After the
ero’s have been used to create a new
canonical form, return to step 3,
using the current canonical form.
Representing simplex tableaus
• Rather than writing each variable in every
constraint, we often used a shorthand
display called a simplex tableau.
• For example,
z+3x1+x2=6
x1+s1=4
2x1+x2+s2=3
z
x1
x2
s1
s2
Basic
rhs variable
1
3
1
0
0
6
Z=6
0
1
0
1
0
4
S1=4
0
2
1
0
1
3
S2=3
Representing simplex tableaus
• This format makes it very easy to
spot basic variable:
Just look for columns having a single
entry of 1 and all other entries equal to
0 (s1 and s2).
In our use of simplex tableaus, we will
encircle the pivot term and denote the
winner of the ratio test by *.
Using the simplex algorithm to
solve minimization problems
• There are two different ways that the
simplex algorithm can be used to solve
minimization problems.
• We illustrate these methods by solving the
following LP:
min z=2x1-3x2
s.t.
x1+ x2≤4
x1- x2≤6
x1,x2≥0
(LP2)
Method 1
• We can find the optimal solution to LP 2 by solving LP 2’:
max -z=-2x1+3x2
s.t.
x1+ x2≤4
x1- x2≤6
x1,x2≥0
(LP2’)
Initial tableau:
-z
x1
x2
s1
s2
rhs
Basic
variable
1
2
-3
0
0
0
-z=0
0
1
①
1
0
4
S1=4
4/1=4*
0
1
-1
0
1
6
S2=6
None
ratio
The ratio test indicates that x2 should enter the basis in row 1.
Optimal tableau
Method 1
-z
x1
x2
s1
s2
rhs
Basic
variable
1
5
0
3
0
12
-z=12
0
1
1
1
0
4
x2=4
0
2
0
1
1
10
S2=10
Thus, the optimal solution to LP 2’ is
-z=12,x2=4,s2=10,x1=s1=0.
then the optimal solution to LP 2 is
z=-12,x2=4,s2=10,x1=s1=0.
Method 1
• In summary, multiply the objective
function for the min problem by -1 and
solve the problem as a maximization
problem with objective function –z.
Note:
(optimal z-value for min problem)
= -(optimal objective function value z for max
problem)
Method 2
• A simplex modification of the ximplex algorithm
can be used to solve min problem directly.
Modify step3 of the simplex as follows:
If all nonbasic variables in row 0 have nonpositive
coefficients, the current bfs is optimal. If any nonbasic
variable in row 0 has a positive coefficient, choose the
variable with the most positive coefficient in row 0 to
enter the basis.
This modification works because increasing a nonbasic
variable with a positive coefficient in row 0 will decrease
z.
Method 2
• To solve LP 2
Initial tableau
Z
x1
x2
s1
s2
rhs
Basic
variable ratio
1
-2
3
0
0
12
z=0
0
1
①
1
0
4
s1=4
4/1=4*
0
1
-1
0
1
10
S2=6
none
Since x2 has the most positive coefficient in row 0, we enter x2
into the basis. the ratio test says that x2 should enter the
basis in row 1.
Method 2
Optimal tableau:
z
x1
x2
s1
s2
rhs
Basic
variable
1
-5
0
-3
0
-12
z=-12
0
1
1
1
0
4
x2=4
0
2
0
1
1
10
S2=10
Thus, the optimal solution to LP 2 is
z=-12,x2=4,s2=10,x1=s1=0.
Alternative optimal solutions
• If an LP has more than one optimal
solution, we say that it has multiple
or alternative optimal solutions.
• Reconsider the Dakota furniture
example, with the modification that
the tables sell for $35 instead of
$30.
Alternative optimal solutions
Initial tableau for Dakota furniture($35/table)
z
x1
x2
x3
s1
s2
s3
s4
Basic
rhs variable
1
-60
-35
-20
0
0
0
0
0
Z=0
0
8
6
1
1
0
0
0
48
S1=48
48/8=6
0
4
2
1.5
0
1
0
0
20
S2=20
20/4=5
0
②
1.5
0.5
0
0
1
0
8
S3=8
8/2=4*
0
0
1
0
0
0
0
1
5
S4=5
None
ratio
The ratio test indicates that x1 should be entered in row 3
Alternative optimal solutions
first tableau for Dakota furniture($35/table)
z
x1
x2
x3
s1
s2
s3
s4
Basic
rhs variable
1
0
10
-5
0
0
30
0
240
Z=240
0
0
0
-1
1
0
-4
0
16
S1=16
None
0
0
-1
0.5
0
1
-2
0
4
S2=4
4/0.5=8*
0
1
0.75 0.25 0
0
0.5
0
4
x1=4
4/0.25=16
0
0
1
0
0
1
5
S4=5
None
0
0
ratio
The ratio test indicates that x3 should be entered in row 2
Alternative optimal solutions
Second (optimal) tableau for Dakota furniture($35/table)
z
x1
x2
x3
s1
s2
s3
s4
Basic
rhs variable
1
0
0
0
0
10
10
0
280
Z=280
0
0
-2
0
1
2
-8
0
24
S1=24
0
0
-2
1
0
2
-4
0
8
x3=8
0
1
1.25
0
0
-0.5 1.5
0
2
x1=2*
0
0
1
0
0
0
1
5
S4=5
0
The optimal solution is s1=24,x3=8 x1=2 s4=5 x2=s2=s3=0
Alternative optimal solutions
• In the optimal tableau, there is a
nonbasic variable, x2, has a zero
coefficient in row 0.
• What happens if we enter x2 into the
basis?
Alternative optimal solutions
Another optimal tableau for Dakota furniture($35/table)
z
x1
x2
x3
s1
s2
s3
s4
Basic
rhs variable
1
0
0
0
0
10
10
0
280
Z=280
0
1.6
0
0
1
1.2
-5.6 0
27.2
S1=27.2
0
1.6
0
1
0
1.2
-1.6 0
11.2 x3=11.2
0
0.8
1
0
0
-0.4 1.2
0
-0.8 0
0
0
0.4
0
-1.2 1
1.6
x2=1.6
3.4 S4=3.4
Alternative optimal solutions
Note: because x2 has a zero coefficient in
the optimal tableau’s row 0, the pivot that
enters x2 into the basis does not change
row 0.
• All variables in new row 0 still have
nonnegative coefficients, thus, the new
tableau is also optimal.
• The pivot has not changed the value of z, an
alternative optimal solution is z=280, s1=27.2,
x3=11.2,x2=1.6,s4=3.4,and x1=s3=s2=0.
Alternative optimal solutions
• Any point on the line segment joining two
optimal extreme points will also be
optimal.
 x1  2
Optimalextremepoint1   x2   0 
 x3  8 
 x1  0 
Optimalext remepoint 2   x2   1.6 
 x3  11.2
Alternative optimal solutions
 x1 
 2
0  2c

 x   c 0  (1  c) 1.6   1.6  1.6c 
 2
 

 

 x3 
8 
11.2 11.2  3.2c 
For 0≤c≤1, all points are optimal.
Unbounded LPs
• There exist points in the feasible
region for which z assumes
arbitrarily large (in max problems) or
arbitrarily small (in min problems)
values.
Example 3
• Breadco Bakeries bakes two kinds of
bread: french and sourdough. Each
loaf of french bread can be sold for
36¢, and each loaf of sourdough
bread for 30¢. A loaf of french
bread requires 1 yeast packet and 6
oz of flour; sourdough requires 1
yeast packet and 5 oz of flour.
Example 3
• At present, Breadco has 5 yeast
packets and 10 oz of flour. Additional
yeast packets can be purchased at 3¢
each, and additional flour at 4¢/oz.
formulate and solve an LP that can be
used to maximize Breadco’s profits.
Example 3
Solution
Define
x1=number of loaves of french bread baked
x2=number of loaves of sourdough bread baked
x3=number of yeast packets purchased
x4=number of ounces of flour purchased
Objective is to maximize z=revenues – costs, where
revenues = 36x1+30x2 and costs=3x3+4x4
Thus, objective function is
max z=36x1+30x2-3x3-4x4
Example 3
• Constraint 1
Number of yeast packages used to bake
bread cannot exceed available yeast plus
purchased yeast.
Available yeast + purchased yeast=5+x3
Then
x1+x2≤5+x3 or x1+x2-x3≤5
Example 3
• Constraint 2
Ounces of flour used to bake breads
cannot exceed available flour plus
purchased flour.
Available flour + purchased flour=10+x4
Then
6x1+5x2≤10+x4 or 6x1+5x2-x4≤10
Example 3
• The LP:
max z=36x1+30x2-3x3-4x4
s.t.
x1+ x2-x3
≤5 (yeast constraint)
6x1+5x2
-x4≤10 (flour constraint)
x1,x2,x3,x4≥0
adding slack variables s1 and s2 to the two
constraints.
Example 3
Initial tableau for Breadco:
z
x1
x2
x3
x4
s1
s2
Basic
rhs variable Ratio
1
-36
-30
3
4
0
0
0
Z=0
0
1
1
-1
0
1
0
5
S1=5
5/1=5
0
⑥
5
0
-1
0
1
10
x3=10
10/6=5/3*
The ratio test indicates that x1 should enter the basis in row 2.
Example 3
first tableau for Breadco:
z
x1
x2
x3
x4
s1
s2
rhs
Basic
variable
1
0
0
3
-2
0
6
60
Z=60
0
0
1/6
-1
1/6
1
-1/6
10/3
S1=10/3
20*
0
1
5/6
0
-1/6
0
1/6
5/3
x1=5/3
none
Ratio
The ratio test indicates that x4 should enter the basis in row 1.
Example 3
second tableau for Breadco:
z
x1
x2
x3
x4
s1
s2
rhs
Basic
variable
1
0
2
-9
0
12
4
100
Z=100
0
0
1
-6
1
6
-1
20
x4=20
None
0
1
1
-1
0
1
0
5
x1=5
None
Ratio
The ratio test fails to indicate the row in which x3 should enter
the basis.
Example 3
• What is happening?
Holding the other nonbasic variables at zero,
x4=20+6x3
x1=5+x3
as x3 is increased, both x4 and x1 increase, and
no matter how large we make x3, the
inequalities x4≥0 and x1≥0 will still be true.
Note: an unbounded LP for a max problem
occurs when a variable with a negative
coefficient in row 0 has a nonpositive
coefficient in each constraint.
Unbounded LPs
• If an LP is unbounded, one will
eventually come to a tableau where
one wants to enter a variable into the
basis, but the ratio test will fail.
• An unbounded LP is usually caused by
an incorrect formulation.
example
In the Breadco example, we allowed Breakco to
pay 3+6(4)=27¢ for the ingredients in a loaf of
french bread and then sell the loaf for 36¢.
Thus, each loaf of french bread earns a profit
of 9¢. Since unlimited purchases of yeast and
flour are allowed, it is clear that our model
allows Breadco to manufacture as much french
bread as it desires, thereby, earning
arbitrarily large profits.
example
• Our formulation of the Breadco example
ignored several aspects of reality.
First, we assumed that demand for Breadco’s
products is unlimited.
Second, we ignored the fact that certain
resources to make bread are in limited supply.
Finally, we made the unrealistic assumption that
unlimited quantities of yeast and flour could be
purchased.